Class 12 Physics Alternating Current | Power associated with Inductor |

**Power associated with Inductor**

- Voltage and current associated with the inductor V = V
_{m}sinωt is I = I_{m}sin(ωt –(∏/2)) respectively. - Instantaneous power p = VI = V
_{m}I_{m}sinωt sin(ωt –(∏/2)) - = - V
_{m}I_{m}sinωt cos ωt ( using formula (sin(A –(∏/2)) = – cosA)) - After simplifying we get,
**p = - ((V**_{m}I_{m}sin2ωt)/(2))- Average Power P
_{L}= (1/T)_{0}^{T}∫ p dt - =(-1/T)
_{0}^{T}∫ (V_{m}I_{m}/2) sin2 ωt dt - = - ( V
_{m}I_{m}/2T)_{0}^{T}∫ sin2 ωt dt - After simplifying we will get ,
**P**._{L}= 0- Therefore average power supplied to an inductor over one complete cycle is 0.

** Problem:- **A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.

__Answer:-__

The inductive reactance,

X_{L} = 2πν L = (2x3.14x50x25x10^{-3}) W

= 7.85Ω

The rms current in the circuit is:- I =(V/X_{L}) =(220V)/(7.85Ω) =28A

** Problem**:-A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V,

50 Hz ac supply. (a) What is the maximum current in the coil? (b) What is the time lag between the voltage maximum and the current maximum?

__Answer:-__

Inductance of the inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of the supply voltage, V = 240 V

Frequency of the supply, ν = 50 Hz

(a) Peak voltage is given as: V_{0}=√2V

=√2 x 240 =339.41V

Angular frequency of the supply, ω = 2 πν

= 2π × 50 = 100 π rad/s

Maximum current in the circuit is given as:

I_{0}= (V_{0})/( √(R)^{2} + ω^{2}L^{2})

=(339.41)/((100)^{2}+(100π)^{2}+(0.50)^{2}) = 1.82A

(b) Equation for voltage is given as:

V = V_{0} cos ωt

Equation for current is given as:

I = I_{0} cos (ωt − Φ)

Where,

Φ = Phase difference between voltage and current At time, t = 0.

V = V_{0}(voltage is maximum)

For ωt − Φ = 0 i.e., at time ,

I = I_{0} (current is maximum)

Hence, the time lag between maximum voltage and maximum current is .( Φ/ω)

Now, phase angle Φ is given by the relation,

tan Φ = (ωL)/(R)

= (2 π x 50x 0.5)/(100)

=1.57

Φ = 57.5^{0} = (57.5 π)/(180)rad

ωt = (57.5 π)/(180)

t =(57.5)/(180 x 2 π x 50)

=3.19x10^{-3} s

=3.2ms

Hence, the time lag between maximum voltage and maximum current is 3.2 ms.

** Problem:- **Obtain the answers (a) to (b) in previous problem if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

** Answer**:-

Inductance of the inductor, L = 0.5 Hz

Resistance of the resistor, R = 100 Ω

Potential of the supply voltages, V = 240 V

Frequency of the supply, ν = 10 kHz = 10^{4}Hz

Angular frequency, ω = 2πν= 2π × 10^{4}rad/s

- Peak voltage, V
_{0}=√2V = 240√2 V

Maximum current , I_{0}= (V_{0})/( √(R)^{2} + ω^{2}L^{2})

=(240√2) / ((100)^{2} + (2 π x 10^{4})^{2} x(0.50)^{2})

=1.1 x 10^{-2 }A

(b) For phase difference Φ, we have the relation:

tan Φ = (ωL)/(R)

= (2 π x 10^{4} x 0.5)/(100) =100 π

Φ =89.82^{0} = (89.82 π)/(180) rad

ω t = (89.82 π)/(180)

t= (89.82 π)/(180x2 πx10^{4})

=25 μ s

It can be observed that I0 is very small in this case. Hence, at high

frequencies, the inductor amounts to an open circuit.

In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.

(Magnetisation and demagnetisation of a magnet)

- Current i through the coil entering at point A increase from zero to a maximum value. Flux lines are set up i.e., the core gets magnetised. With the polarity shown voltage and current are both positive. So their product p is positive.

ENERGY IS ABSORBED FROM THE SOURCE.

- Current in the coil is still positive but is decreasing. The core gets demagnetised and the net flux becomes zero at the end of a half cycle. The voltage v is negative (since di/dt is negative). The product of voltage and current is negative, and ENERGY IS BEING RETURNED TO SOURCE.

(One complete cycle of voltage/current . It is clear that the current lags the voltage.)

- Current i becomes negative i.e., it enters at B and comes out of A. Since the direction of current has changed, the polarity of the magnet changes. The current and voltage are both negative. So their product p is positive. ENERGY IS ABSORBED.

- Current i decreases and reaches its zero value at 4 when core is demagnetised and flux is zero. The voltage is positive but the current is negative. The power is, therefore, negative. ENERGY ABSORBED DURING THE ¼ CYCLE 2-3 IS RETURNED TO THE SOURCE.

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